What is the time complexity of Valid Number?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Valid Number?

This problem uses String. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Valid Number?

Insight 1: The problem is best solved using a Deterministic Finite Automaton (DFA) because the rules for valid numbers can be modeled as a state machine. Insight 2: The transitions between states depend on the current character being processed from the input string (digit, sign, dot, exponent, or invalid).

Which companies ask LeetCode 65. Valid Number in interviews?

This problem is frequently asked at Amazon, Apple, Google, Instacart, LinkedIn and 3 more companies.

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Valid Number - LeetCode 65 Solution

Valid Number - Complete Solution Guide

Valid Number is LeetCode problem 65, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a string s , return whether s is a valid number . For example, all the following are valid numbers: "2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789" , while the following are not valid numbers: "abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53" . Formally, a valid number is defined using one of the following definitions: An integer number followed by an optional exponent . A decimal number followed by an optional exponent . An

Detailed Explanation

The problem asks us to determine whether a given string `s` represents a valid number. The definition of a valid number is somewhat complex, including integers, decimal numbers, and optional exponents. An integer is an optional '+' or '-' sign followed by digits. A decimal number is an optional sign followed by digits then a dot, digits then a dot then digits, or a dot then digits. An exponent consists of 'e' or 'E' followed by an integer. The constraints are that the string length is between 1 and 20, and it only contains digits, English letters, '+', '-', and '.'.

Solution Approach

The provided solutions use a DFA to validate the input string. The DFA is represented by a table (or array of maps in some languages) where each row represents a state, and each column (or map key) represents an input character type. The value in each cell indicates the next state to transition to. The algorithm iterates through the input string, character by character, determining the character type and transitioning to the next state based on the DFA table. Finally, it checks if the final state reached is one of the valid final states.

Step-by-Step Algorithm

Step 1: Define the DFA table. This table represents the state transitions based on the input character type (digit, sign, dot, exponent). Each row is a state, and each column is the input type. The values are the next state to move to.
Step 2: Define the set of final (accepting) states. These are the states that indicate a valid number has been processed.
Step 3: Initialize the current state to the start state (0).
Step 4: Iterate through the input string, character by character.
Step 5: For each character, determine its type (digit, sign, dot, exponent, or invalid).
Step 6: Check if there is a valid transition from the current state for the given character type. If not, the string is invalid; return false.
Step 7: Update the current state to the next state based on the DFA table.
Step 8: After processing the entire string, check if the current state is in the set of final states. If so, the string is valid; return true. Otherwise, return false.

Key Insights

Insight 1: The problem is best solved using a Deterministic Finite Automaton (DFA) because the rules for valid numbers can be modeled as a state machine.
Insight 2: The transitions between states depend on the current character being processed from the input string (digit, sign, dot, exponent, or invalid).
Insight 3: Predefining the allowed transitions for each state simplifies the validation process and ensures all valid number formats are accepted, while invalid formats are rejected.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: Amazon, Apple, Google, Instacart, LinkedIn, Meta, Nutanix, TikTok.

Valid Number | LeetCode 65

Hard

String

Time: O(n)Space: O(1)

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class Solution {
    public boolean isNumber(String s) {
        Map<String, Integer>[] dfa = new Map[8];
        dfa[0] = new HashMap<>();
        dfa[0].put("digit", 2);
        dfa[0].put("sign", 1);
        dfa[0].put("dot", 3);

        dfa[1] = new HashMap<>();
        dfa[1].put("digit", 2);
        dfa[1].put("dot", 3);
        
        dfa[2] = new HashMap<>();
        dfa[2].put("digit", 2);
        dfa[2].put("dot", 4);
        dfa[2].put("exp", 5);

        dfa[3] = new HashMap<>();
        dfa[3].put("digit", 4);

        dfa[4] = new HashMap<>();
        dfa[4].put("digit", 4);
        dfa[4].put("exp", 5);

        dfa[5] = new HashMap<>();
        dfa[5].put("digit", 7);
        dfa[5].put("sign", 6);

        dfa[6] = new HashMap<>();
        dfa[6].put("digit", 7);

        dfa[7] = new HashMap<>();
        dfa[7].put("digit", 7);

        Set<Integer> final_states = new HashSet<>();
        final_states.add(2);
        final_states.add(4);
        final_states.add(7);

        int current_state = 0;

        for (char c : s.toCharArray()) {
            String char_type;
            if (c >= '0' && c <= '9') {
                char_type = "digit";
            } else if (c == '+' || c == '-') {
                char_type = "sign";
            } else if (c == '.') {
                char_type = "dot";
            } else if (c == 'e' || c == 'E') {
                char_type = "exp";
            } else {
                return false;
            }

            if (!dfa[current_state].containsKey(char_type)) {
                return false;
            }

            current_state = dfa[current_state].get(char_type);
        }

        return final_states.contains(current_state);
    }
}

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Problem Info

Description

Problem 65: Valid Number - Detailed explanation and solution approaches.

Topics

String

Companies

AmazonAppleGoogleInstacartLinkedInMetaNutanixTikTok

Stats

Difficulty:Hard

Languages:4

Companies:8

Problem #:65

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Define the DFA table. This table represents the state transitions based on the input character type (digit, sign, dot, exponent). Each row is a state, and each column is the input type. The values are the next state to move to.
2
Step 2: Define the set of final (accepting) states. These are the states that indicate a valid number has been processed.
3
Step 3: Initialize the current state to the start state (0).
4
Step 4: Iterate through the input string, character by character.
5
Step 5: For each character, determine its type (digit, sign, dot, exponent, or invalid).
6
Step 6: Check if there is a valid transition from the current state for the given character type. If not, the string is invalid; return false.
7
Step 7: Update the current state to the next state based on the DFA table.
8
Step 8: After processing the entire string, check if the current state is in the set of final states. If so, the string is valid; return true. Otherwise, return false.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because each character in the input string is processed in constant time. The space complexity is constant because the DFA table and the final states set have a fixed size, regardless of the input string.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Regular Expressions

Regular expressions can be used to match the pattern of a valid number. This approach would involve constructing a complex regular expression that captures all the possible valid number formats.

Time: O(n) in the best case (when the regex engine can quickly determine a mismatch), but can be O(n^2) or even exponential in worst-case scenarios due to backtracking in the regex engine.Space: O(1) to store the regex, but internal workings of regex engines may use linear or quadratic space in certain scenarios.

Trade-offs: Regular expressions can be more concise to write but can be harder to debug and may have performance issues depending on the complexity of the regex and the regex engine implementation. The DFA approach offers more control over the validation process and more predictable performance.

✨

Key Insights

1Insight 1: The problem is best solved using a Deterministic Finite Automaton (DFA) because the rules for valid numbers can be modeled as a state machine.
2Insight 2: The transitions between states depend on the current character being processed from the input string (digit, sign, dot, exponent, or invalid).
3Insight 3: Predefining the allowed transitions for each state simplifies the validation process and ensures all valid number formats are accepted, while invalid formats are rejected.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string (or null in some languages): While constraints specify string length >= 1, code should implicitly handle null termination in C or be given a string of length one or greater in other languages. The DFA will start at state 0. If the string is empty, the final state will be 0. State 0 is not a valid final state so will return false as desired. This case is already implicitly handled in most languages.
Edge case 2: String starting with a dot but no following digits: The provided code in python handles the case such as '.' where if it goes to state 3, it never encounters digits after the dot and thus returns false since state 3 is not an accepting state. Similar behavior is observed for cases such as 'e' or '+'.
Edge case 3: Multiple signs are invalid. The DFA only allows a sign at the beginning or after an exponent. Cases like '--1' or '-+3' are correctly rejected because no transitions are defined for these sequences. Each character must have a valid state transition from the current state.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling all valid number formats, especially edge cases with dots and exponents. For example, forgetting to handle '.1' or '3.'.

✗Common mistake 2: Incorrectly handling signs. Signs are allowed at the beginning of the number and after the exponent symbol, but nowhere else.

✗Common mistake 3: Not accounting for the order of operations (e.g., expecting digits before a dot or an exponent).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Validating user input in forms that require numerical data. This ensures that the data entered by users is in a valid format and can be processed correctly.

🏦Real-world scenario 2: Parsing numerical data from configuration files or data streams. This algorithm can be used to ensure that the data is in a valid format before being used in calculations or other operations.

🎮Real-world scenario 3: Compilers and interpreters for programming languages use similar validation logic to recognize numerical literals within the code.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Valid Number"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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